The probability of obtaining the state \(\ket {\text{}_{y}^{\prime }}\)  is \(C_y / \#X_y\) . Heuristically, the ratio \(C_y / \#X_y\)  is as large as a constant. In other words, there is no reason that \(C_0\)  is much larger or smaller than \(C_y\)  for any other \(y \in \mathbb {F}_2^n\) . We argue, with a little help from algebraic geometry, that this is indeed the case. The function \(f_\mathcal {A}\)  defines a morphism between two affine spaces \(f_\mathcal {A}: \mathbb {A}_{\mathbb {F}_2}^m \rightarrow \mathbb {A}_{\mathbb {F}_2}^n\)  where \(\mathbb {A}_{\mathbb {F}_2}^m = \operatorname{Spec}\mathbb {F}_2[T_1, \dots , T_m]\) , and similarly for \(\mathbb {A}_{\mathbb {F}_2}^n\) . Since both \(\mathbb {A}_{\mathbb {F}_2}^m\)  and \(\mathbb {A}_{\mathbb {F}_2}^m\)  are geometrically irreducible, and since 0 is the generic point of these spaces, it follows from the fibre dimension theorem [1]} that \(\dim f_\mathcal {A}^{-1}(0) = \dim \mathbb {A}_{\mathbb {F}_2}^m - \dim \mathbb {A}_{\mathbb {F}_2}^n = m - n\)Here, the dimension is the geometric dimension not the vector space dimension. We also have that for every \(y \in \mathbb {F}_2^n\) , all irreducible components of \(f_\mathcal {A}^{-1}(y)\)  have dimension at least \(\dim f_\mathcal {A}^{-1}(0)\) . Now, by the Lang-Weil theorem [2]}, [3]}, for any \(y \in \mathbb {F}_2^n\)  it holds that \(\#f_\mathcal {A}^{-1}(y)(\mathbb {F}_2) = O(2^{\dim f_\mathcal {A}^{-1}(y)})\)  where the constant depends on \(f_\mathcal {A}\)  but not on \(y\) . This means the ratio
\( \frac{C_y}{\#X_y} = \frac{\#f_\mathcal {A}^{-1}(y)}{\#f_\mathcal {A}^{-1}(y) + \#f_\mathcal {A}^{-1}(0)} \) 
